From Förberedande kurs i matematik 1

If we use the Pythagorean identity and write \displaystyle \sin ^{2}x as \displaystyle 1-\cos ^{2}x, the whole equation written in terms of \displaystyle \cos x becomes

\displaystyle 2\left( 1-\cos ^{2}x \right)-3\cos x=0

\displaystyle

or, in rearranged form,

\displaystyle 2\cos ^{2}x+3\cos x-2=0

With the equation expressed entirely in terms of \displaystyle \cos x, we can introduce a new unknown variable \displaystyle t=\cos x and solve the equation with respect to \displaystyle t. Expressed in terms of \displaystyle t, the equation is

\displaystyle 2t^{2}+3t-2=0

and this second-degree equation has the solutions \displaystyle t=\frac{1}{2} and \displaystyle t=-2 .

In terms of \displaystyle x, this means that either \displaystyle \cos x=\frac{1}{2} or \displaystyle \text{cos }x=-\text{2}. The first case occurs when

\displaystyle x=\pm \frac{\pi }{3}+2n\pi ( \displaystyle n an arbitrary integer),

whilst the equation \displaystyle \text{cos }x=-\text{2 } has no solutions at all (the values of cosine lie between \displaystyle -\text{1 } and \displaystyle \text{1} ).

The answer is that the equation has the solutions

\displaystyle x=\pm \frac{\pi }{3}+2n\pi ( \displaystyle n an arbitrary integer).